Question

According to recent estimates by the Insurance Corporation of British Columbia (ICBC), 20 percent of all auto insurance claims in BC are fraudulent. Suppose that this estimate is correct,
(a) If 10 auto insurance claims are taken at random, use the binomial distribution to determine the probability that:
(i) at least 6 claims are fraudulent

(ii) at most 3 claims are fraudulent
(iii) anywhere between 5 and 8 (inclusive) claims are fraudulent
(iv) at least 1 claim is fraudulent
(b) If 15 auto insurance are taken at random, what is the probability that between 2 and 6 claims (inclusive) are fraudulent (binomial distribution)? Also using the Poisson approximation to the binomial.

Answer 1

here this is binomial with parameter n=10 and p=0.2

a)

i)

P(X>=6)=1-P(X<=5)=

1-∑x=05 (10Cx)px(q)(10-x) =

0.0064

ii)

P(X<=3)=

∑x=03     (10Cx)px(1−p)(10-x)    =

0.8791

iii)

P(5<=X<=8)=

∑x=58     (nCx)px(1−p)(n-x)    =

0.0328

iv)

P(X>=1)=1-P(X=0)=1-(1-0.2)¹⁰ =1-0.1074 =0.8926

b)

here this is binomial with parameter n=15 and p=0.2

P(2<=X<=6)=

∑x=26     (nCx)px(1−p)(n-x)    =

0.8148

for Poisson approximation ; parameter =np=15*0.2 =3

P(2<=X<=6)=

∑x=26 {e-3*3x/x!}=

0.7673